800 Followers Problem!

Relevant wiki: General Diophantine Equations - Problem Solving

Equation is $m^3+n^3=800 \big(m^2+n^2+m+n \big). \qquad (1)$ First note that $m^3+n^3 \stackrel{5}{\equiv} 0$ . We claim that $m+n \stackrel{5}{\equiv} 0$ . If both of $m$ , $n$ are divisible by $5$ this is true. Let $5 \nmid m$ and $5 \nmid n$ . Then we must have either $m^3 \stackrel{5}{\equiv} 1, n^3 \stackrel{5}{\equiv} 4$ or $m^3 \stackrel{5}{\equiv} 2, n^3 \stackrel{5}{\equiv} 3$ , which in turn give $m \stackrel{5}{\equiv} 1, n \stackrel{5}{\equiv} 4$ or $m \stackrel{5}{\equiv} 3, n \stackrel{5}{\equiv} 2$ respectively. Both of cases lead to $m+n \stackrel{5}{\equiv} 0$ .

On the other hand $m^3+n^3 \equiv 0 \pmod{32}$ . Observe that $m, n$ have same parity. If $m, n$ are both odd, then $m^2-mn+n^2$ is odd and this implies that $m+n \equiv 0 \pmod{32}$ . If $m, n$ are both even, then take $m=2m_1$ and $n=2n_1$ and rewrite the equation $(1)$ as $m_1^3+n_1^3=200(2m_1^2+2n_1^2+m_1+n_1) \qquad(2)$ Therefore, $m_1, n_1$ have same parity. If $m_1, n_1$ are both odd, then $m_1+n_1 \equiv 0 \pmod{16}$ and $m+n \equiv 0 \pmod{32}$ . If $m_1, n_1$ are both even, then let $m_1=2m_2$ and $n_2=2n_2$ and rewrite the equation $(2)$ as $m_2^3+n_2^3=50(4m_2^2+4n_2^2+m_2+n_2) \qquad(3)$ If $m_2, n_2$ are both odd, then $m_2+n_2 \equiv 0 \pmod{4}$ and $m+n \equiv 0 \pmod{16}$ . If $m_1, n_1$ are both even, then let $m_2=2m_3$ and $n_2=2n_3$ and rewrite the equation $(3)$ as $2(m_3^3+n_3^3)=25(8m_3^2+8n_3^2+m_3+n_3)$ Hence $m_3+n_3 \equiv 0 \pmod{2}$ and $m+n \equiv 0 \pmod{16}$ . Therefore we have $m+n \equiv 0 \pmod{16}$ at least.

Using the conclusions $m+n \stackrel{5}{\equiv} 0$ and $m+n \stackrel{16}{\equiv} 0$ , one can write $m+n=80k$ . Equation $(1)$ gives $80k \big(6400k^2-3m(80k-m) \big)=800 \big(6400k^2-2m(80k-m)+80k \big),$ or a quadratic in terms of $m$ : $(3k-20)m^2-80k(3k-20)m+800k(8k^2-80k-1)=0,$ The discriminant is $1600k^2(3k-20)^2-800k(3k-20)(8k^2-80k-1)=800 k(3k-20) \big( -2k^2+40k+1 \big),$ which must be a square. Thus $k(3k-20)\big(-2k^2+40k+1 \big)=2l^2 \qquad (4)$ Look mod $4$ , then $2k^4+3k^2 \stackrel{4}{\equiv} 2l^2$ , which implies that $k \equiv 0 \pmod{4}$ and $l$ is even. Write $k=4K$ and $l=2L$ . Equation $(4)$ becomes $2K(3K-5)\big(-32K^2+160K+1 \big)=L^2 \qquad (5)$ Look mod $5$ , then $3K^4+K^2 \stackrel{5}{\equiv} 2L^2$ , which implies that $K \equiv 0 \pmod{5}$ or $K \equiv 2 \pmod{5}$ . From equation $(5)$ one can get $1<K<6$ , otherwise its LHS will be negative. Thus only possibilities for $K$ are $2, 5$ and its easy too see that only $K=5$ makes $L$ an integer.

Finally $m+n=80k=320K=1600$ . Solving above quadratic in terms of $m$ gives $m=820, 780$ . Therefore there are two solutions $(m, n)=(820, 780), (780, 820),$ and gives the answer as $\boxed{3200}$ .

If we put $N = m+n$ we derive $\begin{aligned} N^3 - 3mnN & = 800(N^2 - 2mn + N) \\ mn & = \frac{N(N^2 - 800N - 800)}{3N - 1600} \end{aligned}$ and hence $m,n$ are the roots of the quadratic $\begin{aligned} X^2 - NX + \frac{N(N^2 - 800N - 800)}{3N - 1600} & = 0 \\ \big(X - \tfrac12N\big)^2 & = \frac{N(3200 + 1600N - N^2)}{4(3N - 1600)} \end{aligned}$ and thus $L = N(3N-1600)(3200 + 1600N - N^2)$ must be a perfect square. For $N$ a positive integer, the expression $L$ is positive when $534 \le N \le 1601$ , and checking shows that $L$ is a perfect square only when $N=1600$ . For this value of $N$ , the roots of the quadratic are $780,820$ , which are integers. Thus there are two solutions of the equation, namely $(780,820)$ and $(820,780)$ , which makes the answer $780+820+820+780 = \boxed{3200}$ .